Mathematics of Flight. Headwinds and Tailwinds

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Everett Elliott
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1 Mathematics of Flight Headwinds and Tailwinds

2 Example: With a tailwind, an aircraft travels 180 miles in 45 minutes. With a headwind, it travels 180 miles in one hour. Find the speed of the aircraft in still air (p).

3 Solution: When you have constant speed, the formula for uniform motion is: distance = rate X time Since knots are nautical miles per hour, the first step would be to change the minutes to hours. 45 minutes x 1 hour = = 3/4 hour 60 minutes In the chart above, p represents the speed of the aircraft in still air; w represents the speed of the wind. Chart Rate X Time in Hours = Distance Tailwind p + w X ¾ hour = ¾(p+w) Headwind p - w X 1 hour = p - w

4 (Note: Because the distance in both situations (headwind and tailwind) is 180 miles, the two equations for distance can be set equal.) 3/4 (p + w) = p w 4[3/4 (p + w) = 4 (p w) 3 (p + w) = 4 (p w) 3p + 3w = 4p - 4w 3w = 4p - 4w - 3p 3w + 4w = p 7w = p

5 The aircraft traveled 180 nautical miles with the tailwind and 180 nautical miles with the head wind. Use either distance from the chart on the previous page equal to 180, substituting 7w for p to find w, the speed of the wind. p - w = 180 3/4(p + w) = 180 7w - w = 180 3/4(7w + w) = 180 6w = 180 3/4 (8w) = 180 6w = 180 6w = w = 180 w = w = 30 The wind speed is 30 knots.

6 To find the speed of the aircraft, use either distance from the chart and substitute 30 for w. p - w = 180 3/4(p + w) = 180 p - 30 = 180 4[3/4(p + 30)] = (180)4 p = (p + 30) = 720 p = 210 3(p + 30) = p + 30 = 240 p = p = 210 The speed of the aircraft is 210 knots.

7 Exercise 1 With a tailwind, an aircraft travels 127 nautical miles in 30 minutes. With a headwind, it travels 107 nautical miles in 30 minutes. Find the speed of the aircraft in still air. When you have constant speed, the formula for uniform motion is: distance = rate X time The first step is to convert minutes to hours. 30 minutes x 1 hour = = 1/2 hour 60 minutes 60 30

8 (p represents the speed of the aircraft in still air; w represents the speed of the wind.) Because the times are the same in this exercise, represent the times in terms of distance and set the algebraic representations equal to each other in the equation. 127 = 107 p + w p - w 127 (p - w) = 107 ( p + w) 127p -127w = 107p + 107w 127p -127w - 107p = 107w 20p - 127w = 107w 20p = 107w + 127w 20p = 234w 20p = 234w p = 11.7w Chart Rate X Time in = Distance Hours Tailwind p + w X 1/2 hour = 127 Headwind p - w X 1/2 hour = 107

9 Substitute p in the chart to find the wind speed. 1/2(p + w) = 127 1/2(11.7w + w) = 127 1/2 (12.7w) = 127 1/2 (12.7w) x 2 = 127 x w = w = w = 20 The wind speed is 20 knots. To find the speed of the aircraft in still air, substitute the wind speed in the chart. 1/2(p + w) = 127 1/2(p + 20) = 127 1/2(p + 20) x 2 = 127 x 2 p + 20 = 254 p = p = 234 The aircraft speed is 234 knots in still air.

10 Exercise 2 With a headwind, an aircraft travels 300 nautical miles in 45 minutes. With a tailwind, the aircraft travels 230 nautical miles in 30 minutes. Find the speed of the aircraft in still air. When you have constant speed, the formula for uniform motion is: distance = rate X time The first step is to convert minutes to hours. 45 minutes x 1 hour = = 3/4 hour 60 minutes Chart Rate X Time in Hours = Distance Tailwind p + w X 1/2 hour = 230 Headwind p - w X 3/4 hour = 300

11 By solving for p or w using the headwind data, we can then substitute the result in the tailwind rate and find the aircraft and wind speeds. 3/4(p - w) = 300 3/4(p - w) x 4 = 300 x 4 3 (p - w) = (p - w) = (p - w) = 400 p = w + 400

12 Using the tailwind data: 1/2(p + w) = 230 1/2(w w) = 230 1/2(2w + 400) = 230 1/2(2w + 400) x 2 = 230 x 2 (2w + 400) = 460 (2w ) = w = 60 2w = w = 30 The wind speed is 30 knots. Solve the above: p = w p = p = 430 The aircraft is traveling 430 knots in still air.

13 Exercise 3 With a tailwind, an aircraft traveling at 400 miles per hour can fly 220 nautical miles in 30 minutes. With a headwind, the aircraft can only fly 180 nautical miles in 30 minutes, What is the speed of the wind? When you have constant speed, the formula for uniform motion is: distance = rate X time The first step is to convert minutes to hours. 30 minutes x 1 hour = = 1/2 hour 60 minutes Chart Rate X Time in = Distance Hours Tailwind w X 1/2 hour = 220 Headwind w X 1/2 hour = 180

14 Exercise 3 Solve for w using the headwind data: 1/2(p - w) = 180 1/2(400 - w) = 180 1/2(400 - w) X 2 = 180 X 2 (400 - w) = 360 (400 - w) + w = w 400 = w = w 40 = w The wind speed is 40 knots.

15 Exercise 3 Solve for w using the tailwind data: 1/2(p + w) = 220 1/2(400 + w) = 220 1/2(400 + w) x 2 = 220 x 2 (400 + w) = 440 (400 + w - 400) = w = 40 The wind speed is 40 knots.

16 More Resources Additional Resources are available online at

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